Lesson 3: Electric
Force
Coulomb's Law
The interaction between charged objects
is a noncontact force which acts over some distance of
separation. Charge, charge and distance. Every electrical
interaction involves a force which highlights the
importance of these three variables. Whether it is a
plastic golf tube attracting paper bits, two likecharged
balloons repelling or a charged styrofoam plate
interacting with electrons in a piece of aluminum, there
is always two charges and a distance between them as the
three critical variables which influence the strength of
the interaction. In this section of Lesson 3, we will
explore the importance of these three variables.
Force as a
Vector Quantity
The electrical force, like all forces, is typically
expressed in units of Newtons. Being a force, the
strength of the electrical interaction is a vector
quantity which has both magnitude and direction. The
direction of the electrical force is dependent upon
whether the charged objects are charged with like charge
or opposite charge and upon their spatial orientation. By
knowing the type of charge on the two objects, the
direction of the force on either one of them can be
determined with a little reasoning. In the diagram below,
objects A and B have like charge causing them to repel
each other. Thus, the force on object A is directed
leftward (away from B) and the force on object B is
directed rightward (away from A). On the other hand,
objects C and D have like charge causing them to attract
each other. Thus, the force on object A is directed
rightward (toward object B) and the force on object B is
directed leftward (toward object A). When it comes to the
electrical force vector, perhaps the best way to
determine the direction of it is to apply the fundamental
rules of charge interaction (opposites attract and
likes repel) using a little reasoning.
Electrical force also has a magnitude
or strength. Like most types of forces, there are a
variety of factors which influence the magnitude of the
electrical force. Two likecharged balloons will repel
each other and the strength of their repulsive force can
be altered by changing three variables. First, the
quantity of charge on one of the balloons will effect the
strength of the repulsive force. The more charged a
balloon is, the greater the repulsive force. Second, the
quantity of charge on the second balloon will effect the
strength of the repulsive force. Gently rub two balloons
with animal fur and they repel a little. Rub the two
balloons vigorously to impart more charge to them, and
they repel a lot. Finally, the distance between the two
balloons will have a significant and noticeable effect
upon the repulsive force. The electrical force is
strongest when the balloons are closest together.
Decreasing the separation distance increases the force.
The magnitude of the force and the distance between the
two balloons is said to be inversely related.
The quantitative expression for the
effect of these three variables on electric force is
known as Coulomb's law. Coulomb's law states that the
electrical force between two charged objects is directly
proportional to the product of the quantity of charge on
the objects and inversely proportional to the square of
the separation distance between the two objects. In
equation form, Coulomb's law can be stated as
where
Q_{1}
represents the quantity of charge on object 1 (in
Coulombs),
Q_{2}
represents the quantity of charge on object 2 (in
Coulombs), and d
represents the distance of separation between the two
objects (in meters). The symbol
k is a
proportionality constant known as the Coulomb's law
constant. The value of this constant is dependent upon
the medium that the charged objects are immersed in. In
the case of air, the value is approximately 9.0 x
10^{9} N • m^{2 }/ C^{2}. If
the charged objects are present in water, the value of
k can be reduced by
as much as a factor of 80. It is worthwhile to point out
that the units on k
are such that when substituted into the equation the
units on charge (Coulombs) and the units on distance
(meters) will be canceled, leaving a Newton as the unit
of force.
The Coulomb's law equation provides an
accurate description of the force between two objects
whenever the objects act as
point charges. A
charged conducting sphere interacts with other charged
objects in a manner that it is as though its charge were
located at its center. While the charge is uniformly
spread across the surface of the sphere, the center of
charge can be considered the center of the sphere. The
sphere acts as a point charge with its excess charge
located at its center. Since Coulomb's law applies to
point charges, the distance
d in the equation is
the distance between the centers of charge for both
objects (not the distance between their nearest
surfaces).
The symbols
Q_{1} and
Q_{2} in the
Coulomb's law equation represent the quantities of charge
on the two interacting objects. Since an object can be
charged positively or negatively, these quantities are
often expressed as "+" or "" values. The sign on the
charge is simply representative of whether the object has
an excess of electrons (a negativelycharged object) or a
shortage of electrons (a positivelycharged object). It
might be tempting to utilize the "+" and "" signs in the
calculations of force. While the practice is not
recommended, there is certainly no harm in doing so. When
using the "+" and "" signs in the calculation of force,
the result will be that a "" value for force is a sign
of an attractive force and a "+" value for force
signifies a repulsive force. Mathematically, the force
value would be found to be positive when
Q_{1} and
Q_{2} are of
like charge  either both "+" or both "". And the force
value would be found to be negative when
Q_{1} and
Q_{2} are of
opposite charge  one is "+" and the other is "". This
is consistent with the concept that oppositelycharged
objects have an attractive interaction and likecharged
objects have a repulsive interaction. In the end, if
you're thinking conceptually (and not merely
mathematically), you would be very able to determine the
nature of the force  attractive or repulsive  without
the use of "+" and "" signs in the equation.
Calculations
Using Coulomb's Law
In physics courses, Coulomb's law is often used as a
type of algebraic recipe to solve physics word problems.
Three such examples are shown here.
Example
A
Suppose that two point charges, each with a
charge of +1.00 Coulomb are separated by a
distance of 1.00 meter. Determine the magnitude
of the electrical force of repulsion between
them.

This is not the most difficult mathematical problem
which could be selected. It certainly was not chose for
its mathematical rigor. The problemsolving strategy
utilized here may seem unnecessary given the simplicity
of the given values. Nonetheless, the strategy will be
used to illustrate its usefulness to any Coulomb's law
problem.
The first step of the strategy is the
identification and listing of known
information in variable form. Here we know the charges of
the two objects
(Q_{1}
and
Q_{2})
and the separation distance between them
(d).
The next step of the strategy involves the listing of the
unknown (or desired) information in variable form. In
this case, the problem requests information about the
force. So
F
is the unknown quantity. The results of the first two
steps are shown in the table below.
Given:
Q_{1} = 1.00 C
Q_{2} = 1.00 C
d = 1.00 m

Find:
F_{elect} = ???

The final step of the strategy
involves substituting known values into the Coulomb's law
equation and using proper algebraic steps to solve for
the unknown information. This step is shown
below.
F_{elect} = k
• Q_{1} •
Q_{2 }/
d^{2}
F_{elect} = (9.0
x 10^{9}
N•m^{2}/C^{2})
• (1.00 C) • (1.00 C) / (1.00 m)^{2}
F_{elect} = 9.0 x
10^{9} N
The force of
repulsion of two +1.00Coulomb charges held 1.00meter
apart is 9 billion Newtons.
This is an incredibly large force which compares in
magnitude to the weight of more than 2000 jetliners.
This problem was chosen
primarily for its conceptual message. Objects simply do
not acquire charges on the order of 1.00 Coulomb. In
fact, more likely Q values are on the order of
10^{9} or possibly 10^{6} Coulombs. For
this reason, a Greek prefix is often used in front of the
Coulomb as a unit of charge. Charge is often expressed in
units of microCoulomb (µC) and nanoCoulomb (nC). If
a problem states the charge in these units, it is
advisable to first convert to Coulombs prior to
substitution into the Coulomb's law equation. The
following unit equivalencies will assist in such
conversions.
1 Coulomb = 10^{6} microCoulomb
1 Coulomb = 10^{9}
nanoCoulomb
The problemsolving strategy used in
Example A included three steps:
 Identify and list known
information in variable form.
 List the unknown (or
desired) information in variable form.
 Substitute known values into
the Coulomb's law equation and using proper algebraic
steps to solve for the unknown information. (In some
case, it might be easier to do the algebra using the
variables and then perform the substitution as the
last step.)
This same problemsolving strategy is demonstrated in
Example B below.
Example
B
Two balloons are charged with an identical
quantity and type of charge: 6.25 nC. They are
held apart at a separation distance of 61.7 cm.
Determine magnitude of the electrical force of
repulsion between them.

The problem states the value of
Q_{1} and
Q_{2}. Since these
values are expressed in units of nanoCoulombs (nC), the
conversion to Coulombs must be made. The problem also
states the separation distance
(d). Since distance is given
in units of centimeters (cm), the conversion to meters
will also be made. These
conversions are required since the units of charge and
distance in the Coulomb's constant are Coulombs and
meters. The unknown quantity is the electrical
force (F).
The results of the first two steps
are shown in the table below.
Given:
Q_{1} = 6.25 nC = 6.25 x
10^{9} C
Q_{2} = 6.25 nC = 6.25 x
10^{9} C
d = 61.7 cm = 0.617 m

Find:
F_{elect} = ???

The final step of the strategy
involves substituting known values into the Coulomb's law
equation and using proper algebraic steps to solve for
the unknown information. This substitution and algebra is
shown below.
F_{elect} = k
• Q_{1} •
Q_{2 }/
d^{2}
F_{elect} = (9.0
x 10^{9}
N•m^{2}/C^{2})
• (6.25 x 10^{9}
C) • (6.25 x
10^{9} C) / (0.617
m)^{2}
F_{elect} = 9.23 x
10^{7} N
Note that the "" sign was dropped
from the Q_{1} and Q_{2} values prior to
substitution into the Coulomb's law equation. As
mentioned above, the use of "+" and "" signs in the
equation would result in a positive force value if
Q_{1} and Q_{2} are likecharged and a
negative force value if Q_{1} and Q_{2}
are oppositely charged. The resulting "+" and "" signs
on F signifies whether the force is attractive (a "" F
value) or repulsive (a "+" F value).
Example
C
Two balloons with charges of +3.37 µC
and 8.21 µC attract each other with a
force of 0.0626 Newtons. Determine the
separation distance between the two
balloons.

The problem states the value of
Q_{1} and
Q_{2}. Since these
values are in units of microCoulombs (µC), the
conversion to Coulombs will be made. The problem also
states the electrical force
(F). The unknown quantity is
the separation distance (d).
The results of the first two steps are shown in the table
below.
Given:
Q_{1} = +3.37 µC =
+3.37 x 10^{6} C
Q_{2} = 8.21 µC = 8.21 x
10^{6} C
F_{elect} = 0.0626 N (use a 
force value since it is repulsive)

Find:
d = ???

As mentioned above, the use of
the "+" and "" signs is optional. However, if they are
used, then they have to be used consistently for the Q
values and the F values. Their use in the equation is
illustrated in this problem.
The final step of the strategy
involves substituting known values into the Coulomb's law
equation and using proper algebraic steps to solve for
the unknown information. In this case, the algebra is
done first and the substitution is performed last. This
algebra and substitution is shown below.
F_{elect} = k
• Q_{1} •
Q_{2 }/
d^{2}
d^{2} •
F_{elect} = k •
Q_{1} •
Q_{2 }
d^{2} = k •
Q_{1} •
Q_{2 }/
F_{elect}
d = SQRT(k •
Q_{1} •
Q_{2 }/
F_{elect}
d = SQRT [(9.0 x
10^{9}
N•m^{2}/C^{2})
• (8.21 x 10^{6}
C) • (+3.37 x
10^{6} C) /
(0.0626 N)]
d = Sqrt [ +3.98
m^{2
}]
d = +1.99
m
Comparing
Electrical and Gravitational Forces
Electrical force and gravitational force are the two
noncontact forces discussed in The
Physics Classroom. Coulomb's law equation for
electrical force bears a strong resemblance to Newton's
equation for universal gravitation.
The two equations have a very similar form. Both
equations show an inverse square relationship between
force and separation distance. And both equations show
that the force is proportional to the product of the
quantity that causes the force  charge in the case of
electrical force and mass in the case of gravitational
force. Yet there are some striking differences between
these two forces. First, a comparison of the
proportionality constants  k versus G  reveals that the
Coulomb's law constant (k) is significantly greater than
Newton's universal gravitation constant (G). Subsequently
a unit of charge will attract a unit of charge with
significantly more force than a unit of mass will attract
a unit of mass. Second, gravitational forces are only
attractive; electrical forces could be either attractive
or repulsive.
The inverse square relationship
between force and distance which is woven into the
equation is common to all noncontact forces. This
relationship highlights the importance of separation
distance when it comes to the electrical force between
charged objects. It is the focus of the next
section of the Lesson 3.
Check
Your Understanding
Use your understanding to answer the following
questions. When finished, depress the mouse on the
"popup menu" to view the answers.
1. The Q in
Coulomb's law equation stands for the _____.
a. mass of a charged object

b. # of excess electrons on the object

c. the current of a charged object

d. the distance between charged
objects

e. charge of a charged object


2. The symbol d in
Coulomb's law equation represents the distance from
___.
a. A to B

b. A to D

c. B to C

d. B to D

e. C to D

f. A to G

g. B to F

h. C to E

3. Determine the electrical force of attraction
between two balloons with separate charges of +3.5 x
10^{8} C and 2.9 x 10^{8} C when
separated a distance of 0.65 m.
4. Determine the electrical force of attraction
between two balloons with opposite charges but the same
quantity of charge of 6.0 x 10^{7} C when
separated a distance of 0.50 m.
5. A balloon has been rubbed with wool to give it a
charge of 1.0 x 10^{6} C. A plastic rod with a
charge of +4.0 x 10^{6} C localized at a given
position is held a distance of 50.0 cm above the balloon.
Determine the electrical force of attraction between the
rod and the balloon.
6. A balloon with a charge of 4.0 µC is held a
distance of 0.70 m from a second balloon having the same
charge. Calculate the magnitude of the repulsive
force.
7. At what distance of separation must two
1.00microCoulomb charges be positioned in order for the
repulsive force between them to be equivalent to the
weight of a 1.00kg mass?
